Water Line Pressure Distribution Over Hills
Water Line Pressure Distribution Over Hills
(OP)
From other threads, I've seen that the minimum liquid flow (in gpm) that will result in a line being liquid full is 10.2(ID)^2.5, so for a 4-inch line it is about 326 gpm or 11,000 bbl/day. A big number, but only 8 ft/s so it is within allowable design conditions.
Now assume a real flow of 1,000 bbl/day (33 gpm) of SG 1.0 water in hilly terrain (say a 100 ft high peak followed by a 300 ft peak followed by a 100 ft peak). At that flow rate, much of the line will not be full, and there won't be any manometer-rebound on the downhill sides of the hills (i.e., liquid will run over the peak and scream down the hill, but without the line being full it won't drag the uphill liquid with it much).
If the line empties into a pond (0 psig), would the pump 1 mile away (call it 2 psi/mile friction) have to:
1. Overcome all three uphills (2 psi+(100+300+100)*0.44 psi/ft=222 psi
2. Overcome the first hill, use the rebound to get 100 ft up the second hill, then add enough pressure to get up the rest of the second hill, then coast over the third (call it 2 psi plus 300 ft/0.44 psi/ft = 134 psi), or
3. Some other number
I have an opinion, but nothing really to base it on.
Now assume a real flow of 1,000 bbl/day (33 gpm) of SG 1.0 water in hilly terrain (say a 100 ft high peak followed by a 300 ft peak followed by a 100 ft peak). At that flow rate, much of the line will not be full, and there won't be any manometer-rebound on the downhill sides of the hills (i.e., liquid will run over the peak and scream down the hill, but without the line being full it won't drag the uphill liquid with it much).
If the line empties into a pond (0 psig), would the pump 1 mile away (call it 2 psi/mile friction) have to:
1. Overcome all three uphills (2 psi+(100+300+100)*0.44 psi/ft=222 psi
2. Overcome the first hill, use the rebound to get 100 ft up the second hill, then add enough pressure to get up the rest of the second hill, then coast over the third (call it 2 psi plus 300 ft/0.44 psi/ft = 134 psi), or
3. Some other number
I have an opinion, but nothing really to base it on.
David Simpson, PE
MuleShoe Engineering
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RE: Water Line Pressure Distribution Over Hills
RE: Water Line Pressure Distribution Over Hills
I am with Artisi on this one. If you imagine the water flowing very slowly and just look at the potential energy then you just have to push it up to the top of the 300ft hill and it will free drain the rest of the way.
regards
Stonecold
RE: Water Line Pressure Distribution Over Hills
RE: Water Line Pressure Distribution Over Hills
That is exactly what I'm not looking for. I'm trying to figure out what the pressures would be without any air-eliminators in the system (the vent pipes on the eliminators have frozen and I'm trying to determine what the pressures should be to compare with what they are).
David
RE: Water Line Pressure Distribution Over Hills
<heads above are in addition to flow losses at any given time.>
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RE: Water Line Pressure Distribution Over Hills
With time the air would gradually be entrained out of the system, but at these low velocities it could take practically forever.
If the first and second downslope sections were changed from 4" to 2" you could ensure that any air would be flushed out straight away, but it would still require a pump capable of 500 ft at 33 gpm at startup. The only advantage would be lower energy consumption once the line was running, and a disadvantage would be a much lower peak flowrate that could be achieved.
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Water Line Pressure Distribution Over Hills
My analysis was:
- the downslope of the first hill would be kept full by the water accumulation in the 300 ft upslope. If there is an air pocket at the top of the first hill, it will be at some pressure equal to the fluid height in the 300 ft upslope (which could often be as much as 300 ft, maybe always 300 ft after initial system loading)
- after initial system loading, the downslope of the 300 ft hill would stay full to 100 ft plus friction (the 200 ft on the downslope would be under some amount of vacuum consistent with the vapor pressure of the water), and any water that makes it over the 300 ft hill would displace fluid that is just about to overtop the last hill.
What confused me is the Flannigan Correlation that says you count all uphills at their full height in determining resistance to flow caused by liquid accumulation in a gas line (500 ft in this case). I guess in a gas line the zero rebound assumption makes sense, but I just don't think it does in a "liquid" line that spends a fair bit of time at rest.
David
RE: Water Line Pressure Distribution Over Hills
If there is some why you can get this condition in your line, you need 500 feet of head. You can set up other conditions such that this might not be true, but assuming that this condition is one of the infinate combinations that can be reached by some liquid/gas/air ratio of flowrates and placement of liquid slugs in your pipeline some day, some time and some where, this combination just might be possible. It is precisely why they say up to 500 feet of head can be required.
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RE: Water Line Pressure Distribution Over Hills
I'm sure we are all in agreement with Artisi that if the breather valves are working then the answer is "300 ft plus friction". At the other extreme we seem to agree that the worst case would be 500 ft plus friction. I would like to argue the case that in steady state the pressure will be in the region of 322 ft plus friction.
If we envisage the situation where water is just starting to flow over the top of the first hill we can imagine the water running down the downslope through the air in the pipe. When the water starts trying to flow up the second upslope we have established a seal at the bottom of the first downslope and the air in this downslope is now trapped. I said in my earlier post that this air becomes pressurized. What I forgot is that as you pressurize air it decreases in volume. As the water tries to flow up the second upslope it will require more and more pressure at its base. This pressure is the combination of the pressure of the trapped air bearing on the water surface, plus the growing column of water compressing the air in the first downslope.
In order to actually calculate the pressures we have to start from the point where we have all the conditions and that is the discharge into the pond. The downslope into the pond will be self draining, making the pressure at the top of the last hill atmospheric. The pressure at the bottom of the last upslope is just 100 ft plus friction. Unfortunately I think in SI units and it is hard for me to do quick calcs in feet of water column, but my rough estimate is that this required 100 ft of pressure will come from a water column of about 84 ft at the bottom of the second downslope plus the air pressure of the trapped air at 20 psia. 20 psi is 46 ft of water, making the pressure at the base of the downlope 84 + 46 = 130 ft. This is absolute and is almost equivalent to the 100 ft gauge we had earlier. As I said, I'm not too good in US Units.
When we are calculating pressure volume relationships with gases it's necessary to work in absolute pressures. Before any water was introduced the air pressure was about 14.7 psia. When the water column at the base of this slope gets to 84 ft the pressure becomes (300 / (300-84)) x 14.7 = 20.4 psia. This assumes a constant slope down the downslope.
This makes the pressure at the top of the second hill 20 psia (46 ft) and the pressure at the bottom of the second upslope 300 + 46 = 346 ft (absolute). A similar analysis of the first downslope shows that the column in this downslope is 87 ft and the trapped air bubble is compressed to 110 psia (254 ft). The pressure at the base of the first upslope is therefore 254 + 100 = 354 ft absolute or 322 ft gauge and this is what I believe is the pressure (plus friction) that the pump will see at its discharge.
One interesting result from this analysis is that there will never be any point where the pressure is below atmospheric and the risk of cavitation is therefore low.
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Water Line Pressure Distribution Over Hills
Without backpressure control from the pond, you can't control column break-aways off the 300 foot hill by increasing pressure. You're only option is to control the flow in from the pump. If a column break happens, the column arriving at the bottom of the valley will increase pressure on any water or air ahead and create a large pressure differential when it arrives at the bottom of the hill. That will tend to blow the water ahead of the air straight out to the pond area as the backpressure is always only 15 psi. Once that happens, the air will be free to backflow to the top of the 300 ft hill, and the large pressure differential appears there which will blow the water out of the first valley.
I believe you must always maintain the liquid seal at the bottom of the 2nd valley and never let in any more than what can immediately flow over the top of the 3rd hill to keep it very well balanced. If you loose the liquid seal, you're on the way to 15 psia at the pump.
Not saying it will happen, but I think it sure might if you're not carefull.
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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
RE: Water Line Pressure Distribution Over Hills
This shows the possible pressure imbalance that will happen if the 2nd downslope filling occurs too fast, or pressure at the top of the hill reaches vapor pressure and column break-away happens. At that point transient pressures, in addition to the density pressures that I have shown, can develop from the column impacting any water ahead of it further accelerating evacuation. You can see that pressures on the 3rd upslope don't balance when calculating from left to right and then from right to left, its 140 psi higher than what backpressre you need to control acceleration, so the 3rd slope water column starts moving outbound rather quickly.
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RE: Water Line Pressure Distribution Over Hills
It would be necessary to have a flow of about 150 gpm in a 4" pipe to develop a syphon and get the column break aways you refer to. This variation is unlikely in a normally sized centrifugal pump installation, but perhaps it is because I have not experienced it that I am not too worried about it.
Harvey
RE: Water Line Pressure Distribution Over Hills
The problem with these boundary conditions is that there's not a good (cheap) general pipeline computer code to simulate air/water transients. You need Olga2000, or Sclumberger Pipesim, or perhaps you have to revert to a CFD model. None of which I'm going to get into today.
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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
RE: Water Line Pressure Distribution Over Hills
The simple and economical solution is to change the pump to one that is capable of producing 300ft + friction total head and to repair/replace/install air release valves at the high points.
RE: Water Line Pressure Distribution Over Hills
zdas,
A back pressure valve (even a manual one for start up purposes) might be a convenient not too costly addition and you might consider adding a few other air valves as well.
Have a look at this guide,
h
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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
RE: Water Line Pressure Distribution Over Hills
Best regards
Morten
RE: Water Line Pressure Distribution Over Hills
http://www.wioa.org.au/conf_papers/98/paper1.htm
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RE: Water Line Pressure Distribution Over Hills
For the "thinkers" in this discussion -- a quick thought off the top of my head - assuming the pump is changed out to one that can produce 300ft + the friction loss - would putting an open standpipe at the highest point - 300ft - coupled with the low flow rates thru the pipeline give a result.
RE: Water Line Pressure Distribution Over Hills
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RE: Water Line Pressure Distribution Over Hills
David
RE: Water Line Pressure Distribution Over Hills
RE: Water Line Pressure Distribution Over Hills
The sentence about double LL's was because I did a web search for Stonel software before remembering that the pipeline software was Stoner. Perhaps Stonel was some instrumentation software package that I no longer use.
RE: Water Line Pressure Distribution Over Hills
My earlier analysis showed that the air trapped in the first downslope would increase the overall head by only 13 ft, but this calculation was based on starting up with an empty pipe and no more air being introduced. In real life where Murphy rules more air would be brought in through entrainment etc and the trapped air bubble would grow with time, increasing the required head from the pump.
MortenA, excellent reference - thanks.
Harvey
RE: Water Line Pressure Distribution Over Hills
http://virtualpipeline.spaces.msn.com
"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
RE: Water Line Pressure Distribution Over Hills
http://virtualpipeline.spaces.msn.com
"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
RE: Water Line Pressure Distribution Over Hills
Best regards
Morten